Skip to main content
Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
모든 교과서Hass 15th EditionCh. 8 - Techniques of Integration문제 8.2.82
8장, 문제 8.2.82

Equations (4) and (5) lead to different formulas for the integral of arctan x:
a. ∫ arctan x dx = x arctan x - ln sec(arctan x) + C [Eq. (4)]
b. ∫ arctan x dx = x arctan x - ln √(1 + x²) + C [Eq. (5)]
Can both integrations be correct? Explain.

검증된 단계별 안내
1
Step 1: Understand that both formulas represent the integral \( \int \arctan x \, dx \) but are expressed differently. Our goal is to verify if these two expressions differ by only a constant, which means they are equivalent up to an additive constant \( C \).
Step 2: Recall the trigonometric identity relating \( \sec(\theta) \) and \( \tan(\theta) \): \[ \sec^2(\theta) = 1 + \tan^2(\theta) \] Since \( \theta = \arctan x \), substitute to get: \[ \sec(\arctan x) = \sqrt{1 + x^2} \]
Step 3: Substitute \( \sec(\arctan x) \) in the first formula with \( \sqrt{1 + x^2} \) from the identity above. This shows that: \[ \ln \sec(\arctan x) = \ln \sqrt{1 + x^2} \]
Step 4: Recognize that the difference between the two integral expressions is just the constant of integration \( C \), since the logarithmic terms are equivalent by the identity. Therefore, both formulas are correct and represent the same family of antiderivatives.
Step 5: Conclude that different-looking expressions for an integral can be equivalent if they differ by a constant, which is always possible due to the arbitrary constant in indefinite integrals.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
10m
도움이 되었나요?

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Integration by Parts

Integration by parts is a technique used to integrate products of functions. It is based on the product rule for differentiation and is expressed as ∫u dv = uv - ∫v du. This method is often applied to integrals involving inverse trigonometric functions like arctan x.
추천 영상:
가이드 코스
06:18
Integration by Parts for Definite Integrals

Inverse Trigonometric Functions and Their Properties

The arctan function is the inverse of the tangent function, mapping real numbers to angles. Understanding its properties, such as its range and how it relates to right triangles, helps simplify expressions involving arctan, especially when converting between trigonometric and algebraic forms.
추천 영상:
06:35
Derivatives of Other Inverse Trigonometric Functions

Logarithmic Identities and Simplification

Logarithmic expressions can often be rewritten using identities like ln(sec(arctan x)) = ln(√(1 + x²)). Recognizing these equivalences is crucial to show that different integral forms are actually the same up to a constant, confirming the correctness of both results.
추천 영상:
7:17
Verifying Trig Equations as Identities
관련 실천
교과서 질문

Evaluate the integrals in Exercises 33–52.

∫ sec⁴(x) tan²(x) dx

27
views
교과서 질문

In Exercises 35–68, use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer.

∫ from 0 to 1 of ((e^(-√x)) / √x dx)

3
views
교과서 질문

Use any method to evaluate the integrals in Exercises 55–66.

∫ x / (x + √(x² + 2)) dx

30
views
교과서 질문

Use the formula ∫ f⁻¹(x) dx = x f⁻¹(x) - ∫ f(y) dy, y = f⁻¹(x)

To evaluate the integrals in Exercises 77-80. Express your answers in terms of x.

∫ log₂ x dx

24
views
교과서 질문

Use reduction formulas to evaluate the integrals in Exercises 41–50.

∫ 8 cos^4(2πt) dt

29
views
교과서 질문

Evaluate the integrals in Exercises 1–22.

∫ cos³(2x) sin⁵(2x) dx

24
views